삼각형 넓이 비의 극한

$\mathrm{\triangle} AOS$에서 $\displaystyle \mathrm{\overline{AS}} = \cos \theta , \mathrm{\overline{OS}} = \sin \theta$이므로 $\displaystyle f ( \theta ) = \frac{1}{2} \cdot \sin \theta \cdot \cos \theta = \frac{1}{4} \sin 2 \theta \cdots \text{㉠}$ $\mathrm{\triangle} OPQ$에서 $\mathrm{\angle} POQ = 2 \theta$이므로 $\displaystyle \mathrm{\overline{PQ}} = \sin 2 \theta$ $\mathrm{\triangle} PQR$에서 $\displaystyle \mathrm{\overline{PR}} = \sin ^{2} 2 \theta$, $\displaystyle \mathrm{\overline{QR}} = \sin 2 \theta \cdot \cos 2 \theta$이므로 $\displaystyle g ( \theta ) = \frac{1}{2} \sin ^{3} 2 \theta \cdot \cos 2 \theta \cdots \text{㉡}$ ㉠, ㉡에서 $\displaystyle \lim\limits _{\theta \rightarrow +0} \frac{\theta ^{2} f ( \theta )}{g ( \theta )} = \lim\limits _{\theta \rightarrow +0} \frac{\frac{1}{4} \cdot \theta ^{2} \cdot \sin 2 \theta}{\frac{1}{2} \cdot \sin ^{3} 2 \theta \cdot \cos 2 \theta}$ $\displaystyle = \lim\limits _{\theta \rightarrow +0} \frac{1}{2} \cdot \frac{\theta ^{2}}{\sin ^{2} 2 \theta} \cdot \frac{1}{\cos 2 \theta}$ $\displaystyle = \lim\limits _{\theta \rightarrow +0} \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{( 2 \theta ) ^{2}}{\sin ^{2} 2 \theta} \cdot \frac{1}{\cos 2 \theta}$ $\displaystyle = \frac{1}{8}$ $\therefore p ^{2} + q ^{2} = 64 + 1 = 65$